You start with 2.5L of a KNO3 stock solution and wish to prepare 10.0 L of 1.5 M KNO3. What percentage (m/v) would the potassium nitrate stock solution need to be if you were to use it all? (MW of KNO3: 101g/mole)

Answer 1

L1 x M1 = L2 x M2
2.5 x M1 = 10.0 x 1.5
M1 = ? = 10*1.5/2.5 = 6.0 M
Now convert 6.0M to % w/v.
M = mols/L = 6.0 mols/L.
1 mol = 101 g, then 6 mols must be 6 mol x 101 g/mol = ?